题解 | #输出单向链表中倒数第k个结点#

#include <iostream>
using namespace std;
struct ListNode{
    int m_nKey;
    ListNode* m_pNext = NULL;
};

int main() {
    int n;
    while (cin >> n) {
        ListNode* head = new ListNode;
        ListNode* curr = head;

        for(int i = 0; i < n; i++){
            ListNode* node = new ListNode;
            int val;
            cin >> val;
            node->m_nKey = val;
            curr->m_pNext = node;
            curr = node;
        }

        int k;
        cin >> k;
        ListNode* fast = head->m_pNext;
        ListNode* slow = head->m_pNext;
        while(k>0) {
            fast = fast->m_pNext;
            k--;
        }

        while(fast!=NULL){
            fast = fast->m_pNext;
            slow = slow->m_pNext;
        }

        cout << slow->m_nKey << endl;
    }

    return 0;
}
// 64 位输出请用 printf("%lld")